1002 B
1002 B
The integration by parts formula is:
\int udv = uv - \int vdu
Deriving the Integration by Parts Formula
\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)
- Integrating both sides, we get:
\int \frac{d}{dx} (f(x)g(x))dx = \int [f'(x)g(x) + f(x)]
- Through the distributive property of integrals,
= \int f'(x)g(x)dx + \int f(x)g'(x)dx
- An integral cancels out an antiderivative, therefore:
f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx
- Moving terms around:
\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx
Now, let u = f(x) and v = g(x), then dv = g'(x)dx and du = f'(x)dx.
Examples
Evaluate the below antiderivative using integration by parts.
\int xe^{2x}dx
- Define
uto be a value you can take the derivative of easily, in this caseu = x. The rest of the integral will be set todv, in this case,dv = e^{2x}dx.u = xdu = \frac{d}{dx}(x)= 1dxdv = e^{2x}dxv = \frac{1}{2}e^{2x}- The antiderivative ofdv.